Calculation for problem #1 of the Final

This notebook containe only the calculations. Reffer to typed solution for a more detailed explanation.

Angular integrals

F = {F0, F1, F2}

Defining the matrix element ⟨(Subscript[m, 1]) = x, 
        (Subscript[m, 2]) = y | ((e^2)/(Subscript[r, 12])) | (Subscript[m, 1]) = y, 
        (Subscript[m, 2]) = w⟩

me[x_, y_, z_, w_] := 
      Sum[F[[l + 1]] 
          Sum[ClebschGordan[{1, 0}, {l, 0}, {1, 0}]^2 
              ClebschGordan[{1, x}, {l, m}, {1, z}] 
              ClebschGordan[{1, w}, {l, m}, {1, y}], {m, -l, l}], {l, 0, 
          2}]

F(l)is the radial integral associated with the radial wave function with angular momentum l in the matrix element. (see below for definition, or in typed solution)

\(TraditionalForm\`\(\^1\) D\) states

Notation: |L, 
              (Subscript[m, L])⟩≡ | 
          (Subscript[m, L])⟩.  Note that all \(TraditionalForm\`\(\(\^1\)\(D\)\)\) states come out the same, as expected by rotation invariance. Also note that the energy shift does not depends on (Subscript[F, 1]), as expected.

\(TraditionalForm\`\(\(\(\(\(|\)\(2\)\)\(⟩\)\) = \(\(|\)\(1\)\), 1\)\(⟩\)\)\)

oneD = me[1, 1, 1, 1]

\(TraditionalForm\`\(\(\(\(\(\(|\)\(1\)\)\(⟩\)\) = \(\((\(\(|\)\(1\)\), 0⟩\)\(+\)\) | 0, 
          1\)\(⟩\)\)\()\)\) 
      1\/√2\)

(1/2) me[1, 0, 1, 0] + (1/2) me[1, 0, 0, 1] + (1/2) 
        me[0, 1, 1, 0] + (1/2) me[0, 1, 0, 1]

\(TraditionalForm\`\(\(\(\(\(\(\(\(\(|\)\(1\)\)\(⟩\)\) = \((\(\(|\)\(1\)\), \(-1\)⟩\) + 2 | 0, 
                0\)\(⟩\)\)\(+\)\) | \(-1\), 
          1\)\(⟩\)\)\()\)\) 
      1\/√6\)

(1/6) me[1, -1, 1, -1] + (4/6) me[0, 0, 0, 0] + (1/6) 
        me[-1, 1, -1, 1] + (2/6) me[1, -1, 0, 0] + (2/6) 
        me[0, 0, 1, -1] + (2/6) me[-1, 1, 0, 0] + (2/6) 
        me[0, 0, -1, 1] + (1/6) me[1, -1, -1, 1] + (1/6) 
        me[-1, 1, 1, -1]

\(TraditionalForm\`\(\(\(\(\(\(|\)\(1\)\)\(⟩\)\) = \(\((\(\(|\)\(-1\)\), 0⟩\)\(+\)\) | 
              0, \(-1\)\)\(⟩\)\)\()\)\) 
      1\/√2\)

(1/2) me[-1, 0, -1, 0] + (1/2) 
        me[-1, 0, 0, -1] + (1/2) 
        me[0, -1, -1, 0] + (1/2) 
        me[0, -1, 0, -1]

\(TraditionalForm\`\(\(\(\(\(|\)\(2\)\)\(⟩\)\) = \(\(|\)\(-1\)\), \(-1\)\)\(⟩\)\)\)

me[-1, -1, -1, -1]

\(TraditionalForm\`\(\^3\) P\) states

\(TraditionalForm\`\(\(\(\(\(\(|\)\(1\)\)\(⟩\)\) = \(\((\(\(|\)\(1\)\), 0⟩\)\(-\)\) | 0, 
          1\)\(⟩\)\)\()\)\) 
      1\/√2\)

threeP = (1/2) me[1, 0, 1, 0] - (1/2) me[1, 0, 0, 1] - (1/2) 
          me[0, 1, 1, 0] + (1/2) me[0, 1, 0, 1]

\(TraditionalForm\`\(\(\(\(\(\(|\)\(1\)\)\(⟩\)\) = \(\((\(\(|\)\(1\)\), \(-1\)⟩\)\(-\)\) | \(-1\), 
          1\)\(⟩\)\)\()\)\) 
      1\/√2\)

(1/2) me[1, -1, 1, -1] - (1/2) 
        me[1, -1, -1, 1] - (1/2) 
        me[-1, 1, 1, -1] + (1/2) 
        me[-1, 1, -1, 1]

\(TraditionalForm\`\(\(\(\(\(\(|\)\(1\)\)\(⟩\)\) = \(\((\(\(|\)\(0\)\), \(-1\)⟩\)\(-\)\) | \(-1\), 
          0\)\(⟩\)\)\()\)\) 
      1\/√2\)

(1/2) me[-1, 0, -1, 0] - (1/2) 
        me[-1, 0, 0, -1] - (1/2) 
        me[0, -1, -1, 0] + (1/2) 
        me[0, -1, 0, -1]

\(TraditionalForm\`\(\^1\) S\) states

\(TraditionalForm\`\(\(\(\(\(\(\(\(\(|\)\(1\)\)\(⟩\)\) = \(\((\(\(|\)\(1\)\), \(-1\)⟩\)\(-\)\) | 0, 
                0\)\(⟩\)\)\(+\)\) | \(-1\), 
          1\)\(⟩\)\)\()\)\) 
      1\/√3\)

oneS = (1/3) me[1, -1, 1, -1] + (1/3) 
          me[0, 0, 0, 0] + (1/3) me[-1, 1, -1, 1] - (1/3) 
          me[1, -1, 0, 0] - (1/3) me[0, 0, 1, -1] - (1/3) 
          me[-1, 1, 0, 0] - (1/3) me[0, 0, -1, 1] + (1/3) 
          me[1, -1, -1, 1] + (1/3) 
          me[-1, 1, 1, -1]

Hund's Rule

ΔE = Simplify[{threeP, oneD, oneS}]

Hund's Rule works-      \(ΔE\_\(\(\^3\)\(P\)\)\)<\(ΔE\_\(\(\^1\) D\)\)<\(ΔE\_\(\(\^1\) S\)\). This is true for any radial wavefunction!                  ☺

Radial integrals

Now we assume the outer electrons are

\(TraditionalForm\`∫d\ \(r\_1\) \(r\_1\^2\) \(∫d\ \(r\_2\) \(r\_2\^2\) \(\(\(R\_21\)(r\_1)\)\^2\) \(\(\(R\_21\)(
                    r\_2)\)\^2\) \(\(r\_ < \)\^l\/\(r\_ > \)\^\(l + 1\)\) 
              e\^2\) = \(\(Z\ e\^2\)\/\(288\ a\_0\)\) \(∫d\ \(ρ\_1\) \(ρ\_1\^2\) \(∫d\ \(ρ\_2\) \(ρ\_2\^2\) \(ρ\_1\^2\) \(e\^\(-ρ\_1\)\) \(ρ\_2\^2\) \(e\^\(-ρ\_2\)\) \(ρ\_ < \)\^l\/\(ρ\_ > \)\^\(l + 1\)\)\)\)

Rint0 = ((Z (e^2))/(288 (Subscript[a, 0]))) 
        Integrate[((Subscript[ρ, 1])^(3 - l)) (E^(-(Subscript[ρ, 1])))*
              Integrate[((Subscript[ρ, 2])^(4 + l)) 
                    (E^(-(Subscript[ρ, 2]))) /. {l -> 0}, {(Subscript[ρ, 2]), 
                  0, (Subscript[ρ, 1])}] /. {l -> 0}, {(Subscript[ρ, 1]), 
            0, ∞}]

Rint2 = ((Z (e^2))/(288 (Subscript[a, 0]))) 
        Integrate[((Subscript[ρ, 1])^(3 - l)) (E^(-(Subscript[ρ, 1])))*
              Integrate[((Subscript[ρ, 2])^(4 + l)) 
                    (E^(-(Subscript[ρ, 2]))) /. {l -> 2}, {(Subscript[ρ, 2]), 
                  0, (Subscript[ρ, 1])}] /. {l -> 2}, {(Subscript[ρ, 1]), 
            0, ∞}]

ΔE /. {F0 -> Rint0, F2 -> Rint2}

N[ΔE /. {F0 -> Rint0, F2 -> Rint2}]

Setting the effective Z to 2, we get the Coulomb repulsion energies in eV:

N[ΔE /. {F0 -> Rint0, 
            F2 -> Rint2}] /. {Z -> 2, e ->  (Sqrt[27.2 eV]), 
        (Subscript[a, 0.])  -> 1}